For time period to be shortest $w_{max}$ the fastest possible rate of rotation of a planet for which gravitational force on the equator just barely provides the centripetal force needed for the rotation.
Let M be mass and R the radius of planet.
$mRw_{max}^2=\large\frac{GMm}{R^2}$
$w_{max}=\sqrt {\large\frac{GM}{R^3}}$
$\qquad= \sqrt {\large\frac{G\bigg(\Large\frac{4}{3} \pi R^3 \rho\bigg)}{R^3}}$
$\qquad=2 \sqrt {\large\frac{\pi G \rho}{3}}$
$\large\frac{2 \pi}{T_{ \Large min}}=2 \sqrt {\large\frac{\pi G \rho}{3}}$
$T_{ min} = \sqrt {\large\frac{3 \pi}{G \rho}}$
Hence d is the correct answer.